Every order topology is regular

An ordered space in the order topology is Hausdorff: if \(x<y\), then either there is \(z\) with \(x<z<y\), in which case the rays \((-\infty,z)\) and \((z,\infty)\) separate \(x\) and \(y\), or there is no point between them, in which case \((-\infty,y)\) and \((x,\infty)\) do so.

We prove regularity. Let \(x\in U\) with \(U\) open. Choose a convex basic open set \(B\) with \[ x\in B\subset U. \] We claim there is a convex open neighborhood \(V\) of \(x\) with \[ \overline V\subset B. \] For each finite endpoint of \(B\), choose an interior cut strictly between that endpoint and \(x\) when such a point exists. If no such point exists, then \(x\) is the immediate successor (or predecessor) of that endpoint, so the corresponding one-sided ray beginning at \(x\) is already open. Doing this independently on the left and right gives a convex basic neighborhood \(V\) whose possible closure endpoints still lie inside \(B\). Thus \(\overline V\subset B\subset U\).

Equivalently, if \(A=X\setminus U\) is closed and \(x\notin A\), then \(V\) and \(X\setminus\overline V\) are disjoint open neighborhoods of \(x\) and \(A\), respectively. Hence every order topology is regular.